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3 years ago

PLEASE I NEED HELP!

Mathematics

2 answers:

N76 [4]3 years ago

4 0

Answer:Complete the statements.

When r = 2, the first expression is . 6

When r = 2, the second expression is 6.

When r = 5, the first expression is 21.

When r = 5, the second expression is .

The expressions are not equivalent. 12

Step-by-step explanation:

i got it right on e d g e n u i t y

Step-by-step explanation:

NARA [144]3 years ago

3 0

Fourth one i think (the way u typed the q is confusing)

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Help please need those answers

icang [17]

We know: The sum of the measures of the angles of a triangle is equal 180°.

We have: m∠A =65°, m∠B = (3x - 10)° and m∠C = (2x)°.

The equation:

65 + (3x - 10) + 2x = 180

(3x + 2x) + (65 - 10) = 180

5x + 55 = 180 subtract 55 from both sides

5x = 125 divide both sides by 5

x = 25

m∠B = (3x - 10)° → m∠B = (3 · 25 - 10)° = (75 - 10)° = 65°

m∠C = (2x)° → m∠C = (2 · 25)° = 50°

<h3>Answer: x = 25, m∠B = 65°, m∠C = 50°</h3>

8 0

3 years ago

Please help this is due today

prisoha [69]

Answer:

x=28°

y=152°

Step-by-step explanation:

x=28°

y=152°

3 0

2 years ago

Write the equation in slope-intercept form. 4x+3y=18

Helen [10]

Answer:

y = -4/3x + 6

Step-by-step explanation:

1. 3y - 4x + 3y = 18 - 3y

2. 4x = -3y + 18

3. 18 - 4x = -3y + 18 - 18

4. -3/ -3y = 4x - 18 /3

5. y = -4/3x + 6

6 0

2 years ago

HELP CALCULUS 25 POINTS

shepuryov [24]

Answer:

A.) Max at x = 6 and Min at x = -6

Step-by-step explanation:

We say that f(x) has a relative (or local) maximum at x=c if f(x)≤f(c) f ( x ) ≤ f ( c ) for every x in some open interval around x=c . We say that f(x) has an absolute (or global) minimum at x=c if f(x)≥f(c) f ( x ) ≥ f ( c ) for every x in the domain we are working on.

4 0

3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

3 0

3 years ago